Problem: Simplify; express your answer in exponential form. Assume $n\neq 0, p\neq 0$. $\dfrac{{(n^{-4})^{2}}}{{(n^{-2}p^{2})^{-4}}}$
Answer: To start, try working on the numerator and the denominator independently. In the numerator, we have ${n^{-4}}$ to the exponent ${2}$ . Now ${-4 \times 2 = -8}$ , so ${(n^{-4})^{2} = n^{-8}}$ In the denominator, we can use the distributive property of exponents. ${(n^{-2}p^{2})^{-4} = (n^{-2})^{-4}(p^{2})^{-4}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(n^{-4})^{2}}}{{(n^{-2}p^{2})^{-4}}} = \dfrac{{n^{-8}}}{{n^{8}p^{-8}}}$ Break up the equation by variable and simplify. $\dfrac{{n^{-8}}}{{n^{8}p^{-8}}} = \dfrac{{n^{-8}}}{{n^{8}}} \cdot \dfrac{{1}}{{p^{-8}}} = n^{{-8} - {8}} \cdot p^{- {(-8)}} = n^{-16}p^{8}$.